what is the wavelength, in nanometers, of light with an energy content of 4.84×10² kj mol⁻¹?\nhint: the unit…

Explanation:

Step1: Convert energy per mole to energy per photon

The energy per mole (E_{mol}=4.84\times 10^{2}\text{ kJ/mol}=4.84\times 10^{5}\text{ J/mol}). Using (E = \frac{E_{mol}}{N_{A}}), where (N_{A}=6.022\times 10^{23}\text{ mol}^{-1}). (E=\frac{4.84\times 10^{5}\text{ J/mol}}{6.022\times 10^{23}\text{ mol}^{-1}}) (E = 8.04\times 10^{-19}\text{ J})

Step2: Use the formula (E=\frac{hc}{\lambda}) to find (\lambda)

We know that (h = 6.626\times 10^{-34}\text{ Js}) and (c=3\times 10^{8}\text{ m/s}). From (E=\frac{hc}{\lambda}), we can solve for (\lambda): (\lambda=\frac{hc}{E}) Substitute the values: (\lambda=\frac{6.626\times 10^{-34}\text{ Js}\times3\times 10^{8}\text{ m/s}}{8.04\times 10^{-19}\text{ J}}) (\lambda=\frac{1.9878\times 10^{-25}\text{ Jm}}{8.04\times 10^{-19}\text{ J}}) (\lambda = 2.47\times 10^{-7}\text{ m})

Step3: Convert meters to nanometers

Since (1\text{ m}=10^{9}\text{ nm}), then (\lambda=(2.47\times 10^{-7}\text{ m})\times10^{9}\text{ nm/m}) (\lambda = 247\text{ nm})

Answer:

247